To solve this problem, we use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n$ is the number of trials (79 clients),
• $k$ is the number of successes (less than 10 clients buying the warranty),
• $p$ is the probability of success for a single trial (20% = 0.2),
• $\binom{n}{k}$ is the binomial coefficient (number of combinations).

We are asked to find the probability of fewer than 10 clients buying the warranty. This is the cumulative probability:

$P(X < 10) = P(X = 0) + P(X = 1) + \dots + P(X = 9)$

Steps:

1. Compute the binomial probabilities for $k = 0, 1, 2, \dots, 9$.
2. Sum these probabilities to get $P(X < 10)$.

Let me compute this. scipy.stats import binom

Parameters

n = 79 # number of trials p = 0.2 # probability of success k_values = range(10) # values from 0 to 9

Compute cumulative probability P(X < 10)

probability_less_than_10 = sum(binom.pmf(k, n, p) for k in k_values) probability_less_than_10The binomial probability that fewer than 10 clients out of 79 will buy your extended warranty, given a 20% probability of success, is approximately 0.0322 (3.22%).

This indicates a low likelihood of observing fewer than 10 successes.

Would you like a step-by-step explanation or further calculations? Here are 5 related questions you might explore:

1. What is the probability of exactly 10 clients purchasing the warranty?
2. How does the probability change if the success rate increases to 30%?
3. What is the expected number of clients buying the warranty in this scenario?
4. Can you compute the variance and standard deviation of this binomial distribution?
5. How would the probability change for a different number of clients?

Tip: For large sample sizes and low probabilities, consider using a Poisson approximation to simplify calculations.